Question

How do you solve by substitution: $3x - 8y = - 7$ and $ - 5x - 6y = 3$ ?

Answer 1

Hint: In the given question, we are given two equations that need to be solved by substitution. In order to do so, we first take any equation and isolate any one variable, let’s say variable $x$. After we have isolated $x$, we substitute that value in the other equation and thus our equation becomes homogeneous and easy to solve. We find the value of the variable $y$ then solve further to find the value of $x$

Complete step-by-step solution:
In this question, we are given two equations which we need to solve to find the values of the respective variables.
We solve the given equations by the method of substitution. The given equations are:
$3x - 8y = - 7$ …………….let this be equation (1)
 $ - 5x - 6y = 3$…………….let this be equation (2)
Now, first we need to express any one equation in terms of $x$. Let us consider equation (1):
$3x - 8y = - 7$
Now in order to isolate the variable $x$, let us add $ + 8y$ to both sides of the equation:
Thus we have: $3x = - 7 + 8y$
Therefore, $x = \dfrac{{ - 7 + 8y}}{3}$
Now we have found the value of $x$, let us substitute this value in equation (2):
Equation (2) is $ - 5x - 6y = 3$
$ \Rightarrow - 5\left( {\dfrac{{ - 7 + 8y}}{3}} \right) - 6y = 3$
On multiply the numerator term and we get
$ \Rightarrow \dfrac{{ + 35 - 40y}}{3} - 6y = 3$
On adding the like terms on the left hand side, we get:
$ \Rightarrow \dfrac{{35 - 40y - 18y}}{3} = 3$
Let us multiply both sides of the equation with $3$ , we get:
$ \Rightarrow 35 - 58y = 9$
On adding $ - 35$ to both sides, we get:
$ \Rightarrow - 58y = - 26$
Dividing both sides with $ - 58$, we get:
$y = \dfrac{{13}}{{29}}$
Now as we have already found out that $x = \dfrac{{ - 7 + 8y}}{3}$
Therefore, $x = \dfrac{{ - 7 + 8\left( {\dfrac{{13}}{{29}}} \right)}}{3}$
$ \Rightarrow x = \dfrac{{\dfrac{{ - 203 + 104}}{{29}}}}{3}$
On simplifying and solving it further, we get:
$ \Rightarrow x = \dfrac{{\dfrac{{ - 99}}{{29}}}}{3} = \dfrac{{ - 99}}{{29}} \times \dfrac{1}{3}$
On simplify the term and we get
$ \Rightarrow x = \dfrac{{ - 33}}{{29}}$

Thus we have our required answer as $x = \dfrac{{ - 33}}{{29}}$ and $y = \dfrac{{13}}{{29}}$

Note: An alternate way of solving this question is through simultaneously solving the equations-
The equations are given as:
$3x - 8y = - 7$ …………….let this be equation (1)
 $ - 5x - 6y = 3$…………….let this be equation (2)
Let us multiply equation (1) with $5$ and equation (2) with $3$ so that the coefficients for the variable $x$ become equal and they can be cancelled out on adding.
Thus our new equations are:
$15x - 40y = - 35$
$ - 15x - 18y = 9$
On adding the above equations, we get:
\[\begin{array}{*{20}{c}}
  {15x - 40y = - 35} \\
  {\underline { - 15x - 18y = 9} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\, - 58y = - 26}
\end{array}\]
Therefore, $y = \dfrac{{26}}{{58}} = \dfrac{{13}}{{29}}$
Substituting this value of $y$, in any one equation we can find the value of $x$
Substituting the value of $y$ in equation (2), we get:
$ - 5x - 6\left( {\dfrac{{13}}{{29}}} \right) = 3$
On multiply the term and we get
$ \Rightarrow - 5x - \dfrac{{78}}{{29}} = 3$
On adding $ + \dfrac{{78}}{{29}}$ to both sides of the equation, we get:
$ \Rightarrow - 5x = 3 + \dfrac{{78}}{{29}} = \dfrac{{87 + 78}}{{29}} = \dfrac{{165}}{{29}}$
On dividing both sides of the equation with $ - 5$, we get:
$x = - \dfrac{{\not{{165}}}}{{\not{{145}}}} = - \dfrac{{33}}{{29}}$
Hence we get the required answer.